Gauss problem, negative Pell's equation and odd graphs (Q2893945)

\textit{H. Lu} [Acta Math. Sin. 28, 756--762 (1985; Zbl 0619.12002)] proved that for \(d = 4k^{2n} + 1\), where \(k>1\) and \(n>1\) are integers, one has \(h(d) \equiv 0 \pmod n\), where \(h(d)\) is the class number of the real quadratic number field \({\mathbb{Q}}(\sqrt{d})\). Substituting here \(k=2\) and \(n = 2^m-1\) the author gets \(n | h(F_{m+1})\), where \(F_{m+1} = 2^{2^{m+1}}+1\). This is the first theorem of the paper.NEWLINENEWLINEThe second theorem asserts that \(h(d)>1\), if \(d\) is squarefree and has at least two prime factors and can be written as \(d = a^2 + b^2\) with \(a \alpha - b \beta = \pm 1\), where \(\alpha, \beta\) are coprime and \(\alpha^2 + \beta^2 = \gamma ^2\) for some integer \(\gamma\).NEWLINENEWLINEHere the author invokes a lemma from the paper \textit{A. Grytczuk, F. Luca} and \textit{M. Wójtowicz} [Proc. Japan Acad., Ser. A 76, No. 6, 91--94 (2000; Zbl 0971.11013)] to conclude that under the assumptions of the theorem the equation \(x^2 -dy^2 = -1\) has a solution. Actually, this conclusion was known to Euler (cf. [\textit{F. Lemmermeyer}, Higher descent on Pell conics. II: Two centuries of missed opportunities. \url{arXiv:math.NT/0311296 v1}, 18 Nov 2003, Prop. 1.2 on p. 2.]). The conclusion \(h(d)>1\) could be strengthened to \(h(d)\) \textit{even} as can be deduced from the factorization of \(d\) and Hasse's characterization of quadratic number fields with odd class numbers quoted by the author.